What is the `int ((cosx)^3)/((sinx)^2) dx` ?

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sciencesolve | Teacher | (Level 3) Educator Emeritus

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You should come up with the substitution such that:

`sin x = t =gt cos x dx = dt `

`sin^2 x = t^2`

`cos^3 x = (1 - sin^2 x)*cos x`

Hence, changing the variable yields:

`int (cos^3 x)/ (sin^2 x) dx = int ((1 - t^2)dt)/t^2`

Hence, using the property of linearity of integral yields:

`int ((1 - t^2)dt)/t^2 = int 1/(t^2) dt - int t^2/t^2 dt`

`int ((1 - t^2)dt)/t^2 = int t^(-2) dt - int dt`

`int ((1 - t^2)dt)/t^2 = (t^(-2+1))/(-2+1) - t + c`

`int ((1 - t^2)dt)/t^2 = -1/t - t + c`

Substituting `sin x`  for `t`  yields:

`int (cos^3 x)/ (sin^2 x) dx = -1/sin x - sin x + c`

Hence, evaluating the trigonometric integral using substitution yields `int (cos^3 x)/ (sin^2 x) dx = -1/sin x - sin x + c.`

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